Comments on: how do political polls calculate confidence intervals? http://www.pence08.com/political-polls/how-do-political-polls-calculate-confidence-intervals Fri, 30 Apr 2010 00:21:59 -0500 http://wordpress.org/?v=2.8.4 hourly 1 By: mathsmanretired http://www.pence08.com/political-polls/how-do-political-polls-calculate-confidence-intervals/comment-page-1#comment-106 mathsmanretired Mon, 15 Feb 2010 07:27:59 +0000 http://www.pence08.com/political-polls/how-do-political-polls-calculate-confidence-intervals#comment-106 Say that p is the true proportion, expressed as a fraction, of those who support political party A. This is what you would like to know but can't know with certainty until there is a real election. You ask a number of people whether they support A and suppose m say yes out of n definite replies. (Ignore don't knows and people who don't want to tell you.) If you did the sampling process many times you would get different values of m for the same n value. This variation would be expected to relate to the Binomial distribution which can be approximated by the Normal distribution with large numbers such as >1000 in political sampling. The mean of the Binomial is np and the standard deviation sqrt[np(1 - p)]. Of course we don't know p but m/n is a good estimate of it if the sample was large. This gives a mean of m and standard deviation of sqrt[m(1 - m/n)]. Confidence intervals are expressed as so many standard deviations usually 2. If n = 2400 and m = 960 say this gives you a standard deviation of 24. So 2s.d = 48 or about 2% either way. Note that you need to 4 times the sample size to halve the confidence interval.<br><b>References : </b><br> Say that p is the true proportion, expressed as a fraction, of those who support political party A. This is what you would like to know but can’t know with certainty until there is a real election.

You ask a number of people whether they support A and suppose m say yes out of n definite replies. (Ignore don’t knows and people who don’t want to tell you.)

If you did the sampling process many times you would get different values of m for the same n value. This variation would be expected to relate to the Binomial distribution which can be approximated by the Normal distribution with large numbers such as >1000 in political sampling.

The mean of the Binomial is np and the standard deviation
sqrt[np(1 - p)]. Of course we don’t know p but m/n is a good estimate of it if the sample was large. This gives a mean of m and standard deviation of sqrt[m(1 - m/n)].

Confidence intervals are expressed as so many standard deviations usually 2. If n = 2400 and m = 960 say this gives you a standard deviation of 24. So 2s.d = 48 or about 2% either way.

Note that you need to 4 times the sample size to halve the confidence interval.
References :

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